The Electric Flux through a surface A is equal to the dot product of the electric field and area vectors E and A. The dot product of two vectors is equal to the product of their respective magnitudes multiplied by the cosine of the angle between them.
In most cases of this type, it is already given in the problem.
Note that the surface area vector is always perpendicular and outward from the surface.
The cosine of the angle between the two vectors multiplied by the electric field vector is equal to the component of electric field that is perpendicular to the surface area vector.
Electric flux is the product of Newtons per Coulomb (E) and meters squared. Proper units for electric flux are Newtons meters squared per coulomb.
Net electric flux through a closed surface with enclosed charge q is the integral of the dot product between the electric field and the instantaneous surface area vector. The integral of the instantaneous surface area is simply the surface area vector. The electric field at a distance outside a Gaussian surface will be constant at that specific distance.
Choose one that best fits its dimensions. A solid sphere or spherical shell of charge Q would require the use of a sphere, while a line or rod of charge would require a cylinder. A solid sphere or hollow spherical shell with uniform charge distribution can be treated as if all charge were concentrated at the center (a point charge), therefore the radius of your Gaussian surface would be the radius of your sphere plus the distance away from the sphere’s surface.
Frequent formulas are 4pi r squared and pi r squared.
With the proper Gaussian surface, the electric field and surface area vectors will nearly always be parallel.
The permittivity constant epsilon zero is equal to 8. 85E-12.
If given charge density, it is possible to solve for the enclosed charge by multiplying the density by the dimensions of the charge distribution (see above formulas). Note that Q total is the same as the total charge enclosed by your Gaussian surface.
